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## 19 January 2012

### A quantitative demonstration with springs... one that DOESN'T work!

All year long, the majority of my in-class time is spent doing quantitative demonstrations, in which the solution to an example problem is verified live in class via measurement.  Each demonstration seems to end the same way: I perform the measurement, we all see that the measurement matches the prediction within 10% or so... everyone exhales, and I say "Physics Works."

It's not a problem, per se, but by November or so the novelty of my approach has worn off.  Instead of bright shiny faces anticipating whether or not the experiment will match the prediction, I start to see resignation:  "Yeah, physics works, I know, we've done this a million times."  I've got to throw a changeup.

Once we've introduced the force and potential energy of a spring, I set up what looks like a routine quantitative demonstration.  I hang a mass from a spring, as shown in the diagram.  I determine the spring constant via the 5-second method that I discuss in this post.  This particular spring (the one on the left in the picture) has a spring constant of about 7 N/m.

Next, I hang 100 g of mass from the spring, and allow the spring to come to rest.  Using F=kx, I derive that the mass should stretch the spring by 14 cm; sure enough, a ruler placed alongside the spring shows 14 cm of stretch.  Physics works.

Now, I use the ruler to measure an additional 5 cm of stretch beyond the equilibrium position.  The question: After I release the mass from rest, what will its speed be when it passes through the equilibrium position again?

[Note to readers before you flame me about these next couple of paragraphs:  Keep reading.  Trust me.]

I use the force on a spring equation, F=kx, to calculate the force on the mass to be (7 N/m)(0.05 m) = 0.35 N.  Then, Fnet = ma, so the acceleration of the mass is (0.35 N)/(0.10 kg) = 3.5 m/s2.

I take a brief interlude, if someone asks, to (correctly) explain that with a vertical spring, it's completely correct and most simple to treat it as if it were a horizontal spring, but with the x = 0 position at the place where the mass would hang at rest.  In other words, it is acceptable to consider that the spring force kx is the net force, as long as we define x = 0 at the resting position.

Now we have enough information to make a kinematics chart.  The mass starts from rest, its acceleration is 3.5 m/s2 and it will move a distance of 5 cm.  Using the kinematics equation v2 = vo2 + 2ax, the speed at the equilibrium position is 0.59 m/s, or 59 cm/s.

"How do we verify this prediction?" I ask the class.  They quickly and accurately tell me to place a motion detector below the mass, and to look at the velocity-time graph.  I do so, I read the v-t graph, and I see that it says about 40 cm/s.  Physics wor------  oops.  Physics DOESN'T work.

All year, I've conditioned the class to expect measurements accurate to 10 or 15%.  I'm not even getting within 30% this time... that's not a matter of "experimental error," especially when I try again and get the same result; or even more especially when I try it with the other spring, and I'm still off by more than 30%.  So, why didn't physics work?

Despite the desperate grasping at straws from some students, someone usually comes up with the right answer:  I used kinematics when acceleration was not constant.  The force of the spring was only 0.35 N immediately after release.  As the spring compressed, the spring force got smaller by F = kx.  So the acceleration got smaller.  So the mass didn't speed up as rapidly, and ended up going less than 59 cm/s.

How do we solve this problem correctly?  Using conservation of energy, of course.  Spring potential energy is converted to kinetic energy, 1/2 kx2 = 1/2 mv2.  Solving, I get v = 41 cm/s.  Now, physics works.

Side note:  Sure, occasionally a student will try to stop me from using kinematics with changing acceleration.  My response is a quick "shhh!!!," Dr. Evil Style, but with a smile.

GCJ